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Homework Help: Maximum Value of gradient

  1. Nov 18, 2006 #1
    when u=-0.6i+0.8j, Du f(1,2)=? gradient f=(18,9)
    I use unit vector dot (-0.6,0.8) dotted with the gradient f, which turns out to be -3.6
    so here is the question. A: the maximum value of Dv f(1,2) is=?
    B: this maximum occurs when v=(?,?)
    I'm confued with the Dv thing. For my guess is that I should use (1,2) to creat a unit vector which is (1/sqrt5, 2/sqrt5), and use that unit vetor dotted with the gradient f, which gives me the maximum value of Dv f(1,2)
    then i will find a vector that its dot product with ( 1/sqrt5, 2/sqrt5) is 0, and tha should be the direction of V which is the answer for part B

    Am I doing this right? plaese help me out. I've been looking into my text book but couldn't find the symbol Dv, they all used Du. Thank u
  2. jcsd
  3. Nov 21, 2006 #2


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    Homework Helper

    Directional Derivatives

    Directional Derivatives

    Directional derivatives of $f(x,y)$ are, for unit vectors [tex]\vec{u}[/tex], are defined by:

    [tex]D_{\vec{u}}f(x,y) = \vec{u}\cdot \vec{\nabla}f (x,y)[/tex]​

    where [tex]\cdot[/tex] denotes the dot product. To see when the maximum value of the directional derivatives occurs, consider the magnitude of the directional derivative, recalling that [tex]\left|\vec{u}\cdot\vec{v} \right| = \left|\vec{u}\right|\cdot \left|\vec{v} \right|\cos \theta[/tex] where [tex]\theta[/tex] is the acute angle between the vectors we have that

    [tex]\left| D_{\vec{u}}f(x,y)\right| = \left| \vec{u}\cdot \vec{\nabla}f (x,y)\right| = \left| \vec{u}\right|\cdot\left| \vec{\nabla}f (x,y)\right|\cos \theta = 1\cdot\left| \vec{\nabla}f (x,y)\right|\cos \theta = \left| \vec{\nabla}f (x,y)\right|\cos \theta[/tex]​

    since [tex]\left| \vec{u}\right| =1[/tex] since [tex]\vec{u}[/tex] is a unit vector; consider that the quantity [tex] \left| \vec{\nabla}f (x,y)\right|\cos \theta[/tex] is maximized when [tex]\cos \theta =1[/tex], i.e. when [tex]\theta =0[/tex] namely when the vectors [tex]\vec{u}[/tex] and [tex]\vec{\nabla}f (x,y)[/tex] are paralell. This occurs if [tex]\vec{u} = \frac{\vec{\nabla}f (x,y)}{\left| \vec{\nabla}f (x,y)\right|}[/tex] which is a unit vector paralell to [tex]\vec{\nabla}f (x,y)[/tex]. The maximum value is then

    [tex]D_{\frac{\vec{\nabla}f (x,y)}{\left| \vec{\nabla}f (x,y)\right|}}f(x,y) = \frac{\vec{\nabla}f (x,y)}{\left| \vec{\nabla}f (x,y)\right|}\cdot \vec{\nabla}f (x,y) = \frac{\left| \vec{\nabla}f (x,y)\right|^{2}}{\left| \vec{\nabla}f (x,y)\right|} = \left| \vec{\nabla}f (x,y)\right|[/tex]​
    Last edited: Nov 21, 2006
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